问题
解答题
计算: (1)(
(2)lg
|
答案
(1)原式=2
×6×31 3
×6+(21 2
×21 2
)1 4
-14 3
=22×33+2
×3 4
-14 3
=108+2-1
=109;
(2)原式=lg(
÷1 2
×12.5)-5 8
×lg9 lg8 lg8 lg27
=lg(
×1 2
×8 5
)-25 2
×2lg3 3lg2 3lg2 3lg3
=lg10-2 3
=1-2 3
=
.1 3